$A$ capacitor of capacitance $C = 10 \ \mu F$ is connected to a $12 \ V$ battery. When a dielectric slab of dielectric constant $K = 5$ is inserted between its plates,how much additional charge (in $\mu C$) will flow from the battery to the capacitor?

The capacitance of a parallel plate capacitor is $2.5 \mu F$. When it is half filled with a dielectric as shown in the figure,its capacitance becomes $5 \mu F$. The dielectric constant of the dielectric is

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_0$ and $W_0$. If the air is replaced by glass (dielectric constant $K = 5$) between the plates,the capacity of the plates and the energy stored in it will respectively be

Between the plates of a parallel plate capacitor, a dielectric plate is introduced to completely fill the space between the plates. The capacitor is charged and then disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference $V$ across the plates versus the length $x$ of the dielectric plate drawn out is:

Half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,then the new capacitance will be:

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then: